1. **State the problem:** Determine on which intervals the function $f(x) = |9 - x^2|$ is decreasing. 2. **Analyze the function:** Inside the absolute value, we have the quadratic $g(x) = 9 - x^2$, which is zero at $x = -3$ and $x = 3$. 3. **Rewrite the function as piecewise:** $$ f(x) = \begin{cases} 9 - x^2 &\text{if } 9 - x^2 \geq 0 \Rightarrow -3 \leq x \leq 3 \\ -(9 - x^2) = x^2 - 9 &\text{if } 9 - x^2 < 0 \Rightarrow x < -3 \text{ or } x > 3 \end{cases} $$ 4. **Find derivatives on each piece:** - For $x \in [-3,3]$, $f(x) = 9 - x^2$, so $$f'(x) = -2x$$ - For $x < -3$ or $x > 3$, $f(x) = x^2 - 9$, so $$f'(x) = 2x$$ 5. **Determine decreasing intervals:** - On $[-3,3]$, $f'(x) = -2x$: - $f'(x) < 0$ when $x > 0$, so $f$ is decreasing on $(0,3)$ - On $(-\infty, -3)$, $f'(x) = 2x$: - Since $x < -3 < 0$, $f'(x) < 0$, so $f$ is decreasing on $(-\infty, -3)$ - On $(3, \infty)$, $f'(x) = 2x$: - Since $x > 3 > 0$, $f'(x) > 0$, so $f$ is increasing on $(3, \infty)$ 6. **Combine the decreasing intervals:** $$(-\infty, -3) \cup (0,3)$$ 7. **Match with the options:** Option A) $] -\infty, -3[ \cup ]0,3[$ matches the result. **Final answer: Option A**